Question:

A rubber contains 70 wt% butadiene (molar mass = 54 g/mol), 20 wt% isoprene (molar mass = 68 g/mol), 5 wt% sulfur and 5 wt% carbon black. Assume that all the sulfur is present in crosslinks. If each sulfide crosslink contains an average of two sulfur atoms, the percentage of possible crosslinks that are joined by vulcanization is ........................%. (Round off to one decimal place)

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In crosslink problems: - First calculate moles of sulfur atoms. - Divide by 2 if each crosslink uses 2 sulfur atoms. - Compare with maximum crosslinking sites = half the total number of monomer units.
Updated On: Aug 27, 2025
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Correct Answer: 9.4

Solution and Explanation

Step 1: Assume a 100 g sample of the rubber.
- Butadiene = 70 g, Isoprene = 20 g, Sulfur = 5 g, Carbon black = 5 g.

Step 2: Moles of each monomer.
\[ n_{\text{butadiene}} = \frac{70}{54} = 1.296 \; \text{mol} \] \[ n_{\text{isoprene}} = \frac{20}{68} = 0.294 \; \text{mol} \] Total moles of monomer units: \[ n_{\text{total}} = 1.296 + 0.294 = 1.59 \; \text{mol} \]

Step 3: Moles of sulfur atoms.
\[ n_S = \frac{5}{32} = 0.156 \; \text{mol of sulfur atoms} \] Each crosslink consumes on average 2 sulfur atoms → \[ n_{\text{crosslinks}} = \frac{0.156}{2} = 0.078 \; \text{mol crosslinks} \]

Step 4: Maximum possible crosslink sites.
- Each monomer unit (butadiene or isoprene) contains one double bond → potential site for crosslinking. - So maximum possible crosslinks ≈ total number of monomer units = 1.59 mol.



Step 5: Fraction of crosslinked sites.
\[ % \text{ crosslinks} = \frac{n_{\text{crosslinks}}}{n_{\text{total}}} \times 100 \] \[ = \frac{0.078}{1.59} \times 100 = 4.9% \] Check carefully: Some formulations take crosslink as connecting two monomers, i.e. each crosslink consumes 2 monomer sites. So maximum crosslinks = \(n_{\text{total}}/2 = 0.795 \; mol\). Thus, \[ % \text{ crosslinks} = \frac{0.078}{0.795} \times 100 = 9.8% \] Since question states “percentage of possible crosslinks” (meaning per two sites), the correct interpretation is second case. Final Answer: \[ \boxed{9.8%} \]
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